We can take the
average interparticle spacing in the gas to be approximately where is the volume and is the number of particles. When the thermal de Broglie wavelength is much smaller than the interparticle distance, the gas can be considered to be a classical or
Maxwell–Boltzmann gas. On the other hand, when the thermal de Broglie wavelength is on the order of or larger than the interparticle distance, quantum effects will dominate and the gas must be treated as a
Fermi gas or a
Bose gas, depending on the nature of the gas particles. The critical temperature is the transition point between these two regimes, and at this critical temperature, the thermal wavelength will be approximately equal to the interparticle distance. That is, the quantum nature of the gas will be evident for \frac{V}{N\lambda_\text{th}^3} \le 1 \ , \text{ or} --> \displaystyle \frac{V}{N\lambda_\text{th}^3} \le 1 \ , \text{ or } \left( \frac{V}{N} \right)^{1/3} \le \lambda_\text{th} i.e., when the interparticle distance is less than the thermal de Broglie wavelength; in this case the gas will obey
Bose–Einstein statistics or
Fermi–Dirac statistics, whichever is appropriate. This is for example the case for electrons in a typical metal at
T = 300
K, where the
electron gas obeys
Fermi–Dirac statistics, or in a
Bose–Einstein condensate. On the other hand, for \frac{V}{N\lambda_\text{th}^3} \gg 1 --> \displaystyle \frac{V}{N\lambda_\text{th}^3} \gg 1 \ , \text{or} \ \left( \frac{V}{N} \right)^{1/3} \gg \lambda_\text{th} i.e., when the interparticle distance is much larger than the thermal de Broglie wavelength, the gas will obey
Maxwell–Boltzmann statistics. Such is the case for molecular or atomic gases at room temperature, and for
thermal neutrons produced by a
neutron source. ==Massive particles==